Incenter Angle

$\angle ABC +\angle ACB = 180^\circ- \angle BAC = 180^\circ- 92^\circ = 88^\circ \\ \implies \angle IBC +\angle ICB = 1/2(\angle ABC+\angle ACB) = 44^\circ \\ \implies \angle BIC = 180^\circ - 44^\circ = 136^\circ \implies \angle EIF = 136^\circ$

Let $\angle ABC = 2b$ and $\angle ACB=2a$ , $2a+2b=180-92=88$ so $a+b=44$ . Thus $\angle BIC=180-(a+b)=180-44=136$ and since angles on opposite of two intersecting straight lines are the same, $\angle BIC = \angle EIF=136$ .

[Latex edits - Calvin]

in triangle ABC, ∠A + ∠B +∠C = 180. given, ∠A= 92. Substituting, we have ∠B + ∠C =88. ∠FCB +∠EBC = 44. also, ∠FCB + ∠EBC +∠BIC = 180. therfore, ∠BIC = 136 =∠EIF. (vertically opposite angles)

Angle BAD = 92/2 = 46 Replace angle ABE with angle x. Hence Angle EBD is also x, and it follows that angle FEC = FCB = 44 - x Angle BEC is 180- x - 2(44 - x) = 92 + x Hence angle EIC is 44 degrees. Angle EIF = 180 - EIC= 136

Angle EIF = Angle BIC Angle ABC + Angle ACB = 88 Deg Angle IBC +Angle ICB = 1/2 of 88 Deg = 44 Deg BIC (and EIF ) = 180 - 44= 136 Deg.

First, label angle ABI and IBD as x, and angle ACI and ICD as y.

Since the sum of the angles in a triangle is 180, x+y=44. So, BIC=136. Solving for AEI and AFI in the same way gives us 88-x, and 88-y respectively, which gives us 92+x+y for EIF. Thus, EIF=136

It is given that $\angle BAC = 92^\circ$

So, $\angle ABC +\angle ACB = 180^\circ - 92^\circ$

$\Rightarrow \angle ABC +\angle ACB = 88^\circ$

$\Rightarrow \frac {\angle ABC +\angle ACB }{2} = (\frac{88}{2})^\circ$

$\Rightarrow \frac {\angle ABC}{2} + \frac {\angle ACB}{2} = 44^\circ$

$\Rightarrow \angle IBC + \angle ICB = 44^\circ$

$\Rightarrow \angle BIC = 180^\circ - 44^\circ$

$\Rightarrow \angle BIC = 136^\circ$

Now, $FC$ and $BE$ are straight line so, $\angle BIC = \angle EIF$

So, $\angle EIF = \boxed {136^\circ}$

Given the triangle $ABC$ and its angle bisectors $AD, BE,$ and $CF$ which intersect at point $I$ , and $∠BAC = 92^\circ$ . We could let $x$ be the measure of $∠ABC$ , $y$ be that of $∠ACB$ , and $z$ , that of $∠EIF$ .

We could use the fact that the sum of the measures of the three interior angles of a triangle is $180^\circ$ . Applying this fact to triangle $ABC$ , we have $m∠BAC + m∠ABC + m∠ACB = 180$ , or, after substitution and simplification,

$(1) x + y = 88$

We can consider triangle $IBC$ , with angles $IBC$ , $ICB$ , and $CIB$ . Since segment $BE$ is an angle bisector, of $∠ABC$ in particular, and segment $BI$ lies on $BE$ , it follows that $BI$ is a bisector of $∠ABC$ . Hence, $BI$ divides $∠ABC$ into congruent angles $ABI$ and $IBC$ , and $m∠IBC$ (the measure of $∠IBC$ ) $= \frac {m∠ABC} {2} = \frac {x} {2}$ .

In a similar manner, we could show that segment $CI$ is an angle bisector of $∠ACB$ , and that $m∠ICB = \frac {m∠ACB} {2} = \frac {y} {2}$ . Using these information and the sum of the interior angles of a triangle, we have $m∠IBC + m∠ICB + m∠CIB = 180$ , or

$(2) \frac {x} {2} + \frac {y} {2} + m∠CIB = 180$ .

Multiplying $(1)$ by $\frac {1} {2}$ , we have $\frac {x} {2} + \frac{y} {2} = 44$ . Substituting this into $(2)$ , we have $44 + m∠CIB = 180$ , or

$(3) m∠CIB = 136$

Finally, since segments $BE$ and $CF$ intersect at $I$ , they create two pairs of vertical angles, and in each pair, the two angles are congruent. One such pair consists of $∠CIB$ and $EIF$ . We have $m∠CIB = 136$ , so the required answer, the measure of $∠EIF$ , is the same as that of $∠CIB$ . Hence, the measure of $∠EIF = 136^\circ$ .

https://lh5.googleusercontent.com/-m3LBkYeybkQ/UaN1YItXyZI/AAAAAAAAAVY/kp_6_7S245c/w353-h358-no/MMM.JPG Here is a picture of the question

In quadrilateral EIFA we already know $angle A$ = $92$ , angle IFA = angle C + angle B/2 similarly angle IEA = angle B + angle C/2 adding up we get A + IFA + IEA = A + B+ C + (B+C)/2 = 180^ + ( 180^ - 92^)/2 = 224^

Therefore angle EIF= 360^ - 224^= 136^

$AFIE$ is a quadrilateral with $\angle EAF=92 ^\circ$ . It satisfies $\angle EAF+ \angle AFI+ \angle FIE+ \angle IEA=360 ^\circ$ , or $\angle IFA+ \angle EIF+ \angle AEI=268 ^\circ$

Now, in triangle $ABE$ we know that $\angle ABE+ \angle BEA+ \angle EAB=180 ^\circ$ , and $\angle EAB=92 ^\circ$ . Since $BE$ is an angle bisector of $\angle ABC$ , then $\angle ABE=\frac {\angle ABC}{2}$ , so $\frac {\angle ABC}{2}+ \angle BEA+ 92 ^\circ=180 ^\circ$ , and $\angle BEA=88 ^\circ - \frac {\angle ABC}{2}$ .

Now, by applying the same logic in triangle $AFC$ , we acquire $\angle AFC=88 ^\circ - \frac {\angle BCA}{2}$ .

Mow we plug in the expressions for $\angle BEA$ and $\angle AFC$ in the angle sum for the quadrilateral (note that $\angle BEA=\angle AEI$ and $\angle AFC=\angle IFA$ ):

$88 ^\circ - \frac {\angle BCA}{2}+ \angle EIF+ 88 ^\circ - \frac {\angle ABC}{2}=268 ^\circ$

$\angle EIF=92^\circ+ \frac {\angle BCA+\angle ABC}{2}$

In triangle $ABC$ , $\angle ABC+ \angle BCA +\angle CAB=180^\circ$ , $\angle BCA+\angle ABC=180^\circ\ -92^\circ=88^\circ$ .

So, $\angle EIF=92^\circ+ \frac {88^\circ}{2}=136^\circ$

Consider triangle $ABC$ . Let $\theta = m\angle ABC$ . Since $BE$ bisects $\theta$ , $\frac{\theta}{2} = m\angle ABE$ . Also, since the sum of the angles in a triangle is $180^\circ$ , $m\angle ACB = 180^\circ-\left(92^\circ+\theta\right) = 88^\circ - \theta$ and $m\angle ACF = \frac{88^\circ-\theta}{2} = 44^\circ-\frac{\theta}{2}$ .

In triangle $ABE$ , $m\angle AEB = 180^\circ - \left(92^\circ+\frac{\theta}{2}\right) = 88^\circ-\frac{\theta}{2}$ . Now consider triangle $ACE$ . $m\angle AFC = 180^\circ - \left(92^\circ-\left(44^\circ-\frac{\theta}{2}\right)\right) = 44^\circ + \frac{\theta}{2}$ .

Finally, consider quadrilateral $AEIF$ . The sum of angles in a quadrilateral is $360^\circ$ , so $m\angle EIF = 360^\circ - \left(92^\circ + 88^\circ - \frac{\theta}{2} + 44^\circ + \frac{\theta}{2}\right) = 136^\circ$

Because $BE$ and $CF$ are angle bisectors, $\angle IBC=b/2$ and $\angle ICB=c/2$ . This makes $\angle BIC = 180-(b+c)/2$ . $b+c$ can be substituted with $180-a$ giving you $90+a/2$ . BIC is equal to EIF because of vertical angles so $90+92/2=136$ .

Consider the special case of ABC isosceles. Then $\angle ACF=22^\circ$ and $\angle AFC=\angle AEF=66^\circ$ So $\angle EIF=(360-92-66-66)^\circ$ .

Since the statement of the problem implies that the solution is independent of the ratio of the lengths AB and AC, we conclude that $\angle EIF=136^\circ$ always.

We double check the implication to make sure the problem is well-posed: Suppose $\angle ACB=(44-x)^\circ$ then $\angle ABC=(44+x)^\circ$ and $\angle AFC=(66+x/2)^\circ$ , $\angle AEB=(66-x/2)^\circ$ , and as before: $\angle EIF=(360-92-66-66)^\circ$

Let $\angle ABC = \beta, \angle ACB = \gamma$ so $\beta + \gamma$ $=\angle ABC + \angle ACB$ $= 180^\circ - \angle BAC$ $= 180^\circ - 92 ^\circ$ $= 88 ^\circ$ . Since $BI$ and $CI$ are angle bisectors, we have $\angle IBC = \frac {\beta}{2}, \angle ICB = \frac {\gamma}{2}$ . Also, $\angle BIC$ and $\angle EIF$ are opposite angles, implying $\angle BIC = \angle EIF$ . Hence

$\angle EIF = \angle BIC = 180^\circ - \angle IBC - \angle ICB = 180^\circ - \frac {\beta} {2} - \frac {\gamma} {2} = 136 ^\circ.$

let we make isosless triangle ABC AB=AC, if angle <BAC = 92 then <ABC and <ACB will be equal to 44, 44. now draw the lines AD, BE, CF such that thay are angle bisectror of triangle ABC. now we take the triangle AEB, in this triangle angle <ABE is half of the angle <ABC which is 22(<ABE=22). angle <BAE=92 (Which is equal to <BAC) now angle <BEA=180-22-92=66(<BEA=66) Now take triangle AEI the angle <AIE = 180 - <IAE - <IEA =180 - (92/2) - 66 = 68 Now our desier angle is <EIF which is double of (<AIE = 68) so <EIF=2*68=136

In this problem it would be safe to assume that the given triangle ABC is isosceles, with angles ABC and ACB both equal to 44 degrees. (This is consistent with the required angle sum of 180 degrees, if the figure is to be a triangle.)

Since EB and FC are angle bisectors of ABC and ACB, respectively, we can conclude that angles EBC and FCB both measure 22 degrees.

Next, draw a line parallel to BC through the incenter I. Let us call this segment GH, with point G lying on AB and point H lying on AC. We can see that FC and EB are both transversals of the parallel lines BC and GH. This implies that angle EIH is congruent to angle EBC, and similarly, angle FIG is congruent to angle FCB. Therefore, both EIH and FIG measure 22 degrees as well.

Lastly, as GIH can be considered as a 180-degree angle, we can conclude that the total measures of FIG, EIF and EIH is also 180 degrees. Algebraic manipulation will yield the measure of angle EIF to be 136 degrees.

Apologies for not using mathematical formatting and for the haphazard method. I am at work right now as I type this solution, and this is my first time anyway. Thank you for reading! :)

as angle BAC is divided in half, angle BAD is 46 degrees. As bisectors are perpendicular to opposite sides, angle BDA is 90 degrees and angle ABD is 44 degrees. Therefore angle EBD is 22 degrees, and angle BID is 68 degrees. Simply multiplying the angle by 2 gives you 136 degrees for BIC, which is same as angle FIE.

WLOG, we can take the triangle to be isosceles, so that $$\angle ABC=\angle ACB = \frac{180-92}2^\circ=44^\circ \implies \angle ABE=22^\circ, $$

So, in $$\triangle ABE, \angle AEB=(180-92-22)^\circ=66^\circ$$

So, in $$\triangle AIE, \angle AIE=(180-46-66)^\circ=68^\circ$$

Similarly, $$\angle AIF=68^\circ$$

$$\implies \angle EIF =2\cdot 68^\circ=136^\circ $$